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Let $G=PSU_3(q)$ and $q=p^n$, where $n$ is odd. Can we conclude that $PSU_3(p)$ is a subgroup of $G$?

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    $\begingroup$ By the way, you haven't reacted to my answer to your question yesterday about ${\rm PSU}_n(q)$. $\endgroup$
    – Derek Holt
    Commented Sep 22, 2013 at 11:09

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Yes, because $SU_3(q) \subset SU_3(q^2)$ is the subgroup of elements Galois-conjugate to their inverse. $SU_3(p)$ is the subgroup of elements defined over $p$ and Galois-conjugate to their inverse. Since the relevant Galois action is the same, they are the same. Then we mod out by the centers, which preserves inclusions.

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